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Question

Given ECr3+/Cr=0.72,EFe2+/Fe=0.42 V
The potential for the cell, Cr|Cr3+(0.1M)||Fe2+(0.01M)|Fe is:

A
0.26 V
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B
0.26 V
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C
0.339 V
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D
0.339 V
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Solution

The correct option is B 0.26 V
Ecell=0.42(0.72)=+0.30 V
2Cr(s)+3Fe2+(0.01M)2Cr3+(0.1M)+3Fe(s)
Q=[Cr3+]2[Fe2+]3=[0.1]2[0.01]2=104
According to Nernst equation,
E=E0.059nlog10Q
=0.300.0596log104(b=6)
=0.261 V.

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